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Author: Brenda Gunderson, Ph.D., 2015
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Stat 250 Gunderson Lecture Notes
3: Probability
Chance favors prepared minds. -- Louis Pasteur
Many decisions that we make involve uncertainty and the evaluation of probabilities.
Interpretations of Probability
Example: Roll a fair die  possible outcomes = {
}
Before you roll the die do you know which one will occur?
What is the probability that the outcome will be a ‘4’? __________ Why?
A few ways to think about PROBABILITY:
(1) Personal or Subjective Probability
P(A) = the degree to which a given individual believes that the event A will happen.
(2) Long term relative frequency
P(A) = proportion of times ‘A’ occurs if the random experiment (circumstance) is repeated
many, many times.
(3) Basket Model
P(A) = proportion of balls in the basket
that have an ‘A’ on them.
10 balls in the basket: 3 blue and 7 white
One ball will be selected at random.
What is P(blue)? ______________
Note: A probability statement IS NOT a statement about
It IS a statement about
.
.
Discover Basic Rules for Finding Probability through an Example
There is a lot you can learn about probability. One basic rule to always keep in mind is that the
probability of any outcome is always between 0 and 1. Now, there are entire courses devoted
just to studying probability. But this is a Statistics class. So rather than start with a list of
definitions and formulas for finding probabilities, let’s just do it through an example so you can
see what ideas about probability we need to know for doing statistics.
Example: Shopping Online
Many Internet users shop online. Consider a population of 1000 customers that shopped
online at a particular website during the past holiday season and their results regarding
whether or not they were satisfied with the experience and whether or not they received the
products on time. These results are summarized below in table form. Using the idea of
probability as a proportion, try answering the following questions.
On Time
Not On Time
Total
Satisfied
800
20
820
Not Satisfied
80
100
180
Total
880
120
1000
a. What is the probability that a randomly selected customer was satisfied with the
experience?
b. What is the probability that a randomly selected customer was not satisfied with the
experience?
c. What is the probability that a randomly selected customer was both satisfied and received
the product on time?
d. What is the probability that a randomly selected customer was either satisfied or received
the product on time?
e. Given that a customer did receive the product on time, what is the probability that the
customer was satisfied with the experience?
f. Given that a customer did not receive the product on time, what is the probability that the
customer was satisfied with the experience?
Note: We stated the above 1000 customers represented a population. If results were based on a sample that is
representative of a larger population, then the observed sample proportions would be used as approximate
probabilities for a randomly selected person from the larger population.
Great job! You just computed probabilities using many of the basic probability rules or formulas
summarized below and also found in your textbook.

Complement rule
P( AC )  1  P( A)

Addition rule
P( A or B)  P( A)  P( B)  P( A and B)

Multiplication rule
P( A and B)  P( A) P( B | A)

Conditional Probability
P( A | B) 
P( A and B)
P( B)
You did not need the formulas themselves but instead used intuition and approaching it as
some type of proportion. Let’s see how your intuition and the above formulas really do
connect.
In part b you found the probability of “NOT being satisfied”, which is the complement of the
event “being satisfied”, so the answer to part b is the complement of the probability you found
in part a.
In part c, there was a key word of “AND” in
the question being asked. The “AND” is just
the intersection, or the overlapping part; the
outcomes that are in common. The picture at
the right show an intersection between the
event A and B. In a table, the counts that are
in the middle are the “AND” counts; there
were 800 (out of the 1000 customers) that
were both satisfied AND received the product on time. There is a multiplication formula above
for finding probabilities of the AND or intersection of two events, but we did not even need to
apply it; as a table presentation of counts provides “AND” counts directly.
In part d, there was a key word of “OR” in the question being asked. The “OR” is union, the
outcomes that are in either one or the other (including those that are in both). The picture at
the right show an union between the event A and B. Notice that if you start with all of the
outcomes that are in A and then add all of the outcomes that are in B, you have double counted
the outcomes that are in the overlap. So the addition formula above shows you need to
subtract off the intersecting probability once to
correct for the double counting. From the
table, you could either add up the separate
counts of 800 + 20 + 80; or start with the 820
that were satisfied and add the 880 that
received it on time and then subtract the 800
that were in both sets; to get the 900 in all that
were either satisfied or received the product
on time.
Finally, parts e and f were both conditional probabilities. In part e you were first told to
consider only the 880 customers that received the product on time, and out of these find the
probability (or proportion) that were satisfied. There were 800 out of the 880 that were
satisfied. The picture below shows the idea of a conditional probability formula above for P(A |
B), read as the probability of A given B has occurred. If we know B has occurred, then only look
at those items in the event B. The event B, shaded at the right, is our new ‘base’ (and thus is in
the denominator of the formula). Now out of
those items in B, we want to find the
probability of A. The only items in A that are
on the set B are those in the overlap or
intersection. So the conditional formula above
shows you count up those in the “A and B” and
divide it by the base of “B”.
Try It! Go back to parts a to f and add the corresponding shorthand probability notation of what
you actually found; e.g. P(satisfied), P(satisfied | on time) next to each answer.
Now there are a couple of useful situations that can make computing probabilities easier.
Definition:
Two events A, B are Mutually Exclusive (or Disjoint) if ... they do not contain any
of the same outcomes. So their intersection is empty.
We can easily picture disjoint events because the definition is a property about the sets
themselves.
If A, B are disjoint, then P(A and B) = 0. If there are no items in the overlapping part, then man
of the probability results will simplify. For example, the additional rule for disjoint events P(A or
B) = P(A) + P(B).
Another important situation in statistics occurs when the two events turn out to be
independent.
Definition:
Two events A, B are said to be independent if knowing that one will occur
(or has occurred) does not change the probability that the other occurs.
In notation this can be expressed as P(A|B) = P(A).
This expression P(A|B) = P(A) tells us that knowing the event B occurred does not change the
probability of the event A happening.
Now it works the other way around too, if A and B are independent events, then P(B|A) = P(B).
As a result of this independence definition, we could show that the multiplication rule for
independent events reduces to P(A and B) = P(A)P(B).
Finally, this rule can also be extended. If three events A, B, C are all independent then P(A and
B and C) = P(A)P(B)P(C).
So let’s apply these two new concepts to our online shopping example.
Back to the Shopping Online Example
Below are results for a population of 1000 customers that shopped online at a particular
website during the past holiday season. Recorded was whether or not they were satisfied with
the experience and whether or not they received the products on time.
Satisfied
Not Satisfied
Total
On Time
800
80
880
Not On Time
20
100
120
Total
820
180
1000
g. Are being satisfied with the experience and receiving the product on time mutually
exclusive (disjoint)? Provide support for your answer.
h. Are being satisfied with the experience and receiving the product on time statistically
independent? Provide support for your answer.
Hint: go back and compare your answers to parts a, e, and f.
Try It! Elderly People
ElderlyPeople
Suppose that in a certain country, 10% of
the elderly people have diabetes. It is also
known that 30% of the elderly people are
living below poverty level and 5% of the
elderly population falls into both of these
categories.
At the right is a diagram for these events.
Do the probabilities make sense to you?
diabetes
below
poverty
0.05
0.05
0.25
diabetes & below poverty
Neither diabetes
nor below poverty
0.65
a. What is the probability that a randomly selected elderly person is not diabetic?
b. What is the probability that a randomly selected elderly person is either diabetic or living
below poverty level?
c. Given a randomly selected elderly person is living below poverty level, what is the
probability that he or she has diabetes?
d. Since knowing an elderly person lives below the poverty level (circle one)
DOES DOES NOT
change the probability that they are diabetic, the two
events of living below the poverty level and being diabetic (circle one)
ARE
ARE NOT
independent.
In the next example, you are not asked to determine if two events are independent, but rather
put independence to use.
Try It! Blood Type
About 1/3 of all adults in the United States have type O+ blood. Suppose three adults will be
randomly selected.
Hint: randomly selected implies the results should be_______________________.
What is the probability that the first selected adult will have type O+ blood?
What is the probability that the second selected adult will have type O+ blood?
What is the probability that all three will have type O+ blood?
What is the probability that none of the three will have type O+ blood?
What is the probability that at least one will have type O+ blood?
Some final notes…
I. Sampling with and without Replacement
Definitions:
A sample is drawn with replacement if individuals are returned to the eligible pool
for each selection. A sample is drawn without replacement if sampled individuals
are not eligible for subsequent selection.
If sampling is done with replacement, the Extension of Rule 3b holds. If sampling is done
without replacement, probability calculations can be more complicated because the
probabilities of possible outcomes at any specific time in the sequence are conditional on
previous outcomes.
If a sample is drawn from a very large population, the distinction between sampling with and
without replacement becomes unimportant. In most polls, individuals are drawn without
replacement, but the analysis of the results is done as if they were drawn with replacement.
The consequences of making this simplifying assumption are negligible.
II. Sometimes students confuse the mutually exclusive with independence.
Check the definitions.
 The definition for two events to be disjoint (mutually exclusive) was based on a SET
property.
 The definition for two events to be independent is based on a PROBABILITY property.
You need to check if these definitions hold when asked to assess if two events are disjoint,
or if two events are independent.
Mutually Exclusive
Independence
If two events are mutually exclusive then we know that P(A and B) = 0.
This also implies that P(A|B) is equal to 0 (if the two events are disjoint and B did occur,
then the chance of A occurring is 0).
So P(A|B) (which is 0) will not be equal to P(A) if the events are disjoint.
III. Probability rules summary
Below is a summary of the key probability results you need to understand and be able to
use.

Complement rule

Mutually Exclusive (disjoint) Events:
The events A, B are disjoint if “A and B” is the empty set.
Thus, P(A and B) = 0.

Addition Rule (general) P( A or B)  P( A)  P( B)  P( A and B)
If A, B are disjoint, we have P( A or B)  P( A)  P( B)
P( A and B)
Conditional Probability (general)
P( A | B) 
P( B)
Independent Events:
The events A, B are independent if P( A | B)  P( A)
Equivalently, the events A, B are independent
if P( A and B)  P( A) P( B)


P( A C )  1  P( A)
The Stats 250 formula card provides a more extensive list, but remember, you may not need
them as you discovered in your first probability example with the online customers.
Additional Notes
A place to … jot down questions you may have and ask during office hours, take a few extra notes, write
out an extra problem or summary completed in lecture, create your own summary about these
concepts.